1. **State the problem:** We want to find the limit $$\lim_{n \to \infty} n\left(\sqrt{2n^2+1} - \sqrt{2n^2-1}\right).$$ 2. **Use the conjugate to simplify:** To simplify the expression inside the parentheses, multiply and divide by the conjugate: $$n \left(\sqrt{2n^2+1} - \sqrt{2n^2-1}\right) \cdot \frac{\sqrt{2n^2+1} + \sqrt{2n^2-1}}{\sqrt{2n^2+1} + \sqrt{2n^2-1}}.$$ 3. **Apply difference of squares:** This gives $$n \cdot \frac{(2n^2+1) - (2n^2-1)}{\sqrt{2n^2+1} + \sqrt{2n^2-1}} = n \cdot \frac{2}{\sqrt{2n^2+1} + \sqrt{2n^2-1}}.$$ 4. **Simplify numerator and denominator:** The numerator is 2n. The denominator is $$\sqrt{2n^2+1} + \sqrt{2n^2-1}.$$ 5. **Factor out $n$ inside the square roots:** $$\sqrt{2n^2+1} = \sqrt{n^2(2 + \frac{1}{n^2})} = n\sqrt{2 + \frac{1}{n^2}},$$ $$\sqrt{2n^2-1} = n\sqrt{2 - \frac{1}{n^2}}.$$ 6. **Rewrite the denominator:** $$n\left(\sqrt{2 + \frac{1}{n^2}} + \sqrt{2 - \frac{1}{n^2}}\right).$$ 7. **Substitute back:** The expression becomes $$\frac{2n}{n\left(\sqrt{2 + \frac{1}{n^2}} + \sqrt{2 - \frac{1}{n^2}}\right)} = \frac{2}{\sqrt{2 + \frac{1}{n^2}} + \sqrt{2 - \frac{1}{n^2}}}.$$ 8. **Take the limit as $n \to \infty$:** Since $\frac{1}{n^2} \to 0$, we get $$\lim_{n \to \infty} \frac{2}{\sqrt{2 + 0} + \sqrt{2 - 0}} = \frac{2}{\sqrt{2} + \sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}.$$ 9. **Final answer:** $$\boxed{\frac{1}{\sqrt{2}}}.$$